6.5 Trig. Form of a Complex Number
22 babiaz +=+=
Ex. Z = -2 + 5i ( )5,2−
22 5)2( +−=z
29=z -2
5
Trigonometric Form of a Complex Number
( )θθ sincos irz +=r = the hypotenuse and theta = the angle.
Ex. iz 322−−= Find r (the hyp) & theta
r = 4
===2
32
.
.tan
adj
oppθ
1
3
060=∴θ ref. angle
°=∴ 2403
4or
πθ
)3
4sin
3
4(cos4
ππiz +=
Write in trig. form
2−
32−
Write the complex number in standard form a + bi.
Ex. ⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛−+⎟⎠
⎞⎜⎝
⎛−=3
sin3
cos8ππ
iz
⎥⎦
⎤⎢⎣
⎡−= iz
2
3
2
122
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z = 2 − 6i
Multiplication of Complex Numbers
( ) ( )[ ]21212121 sincos θθθθ +++= irrzz
Find the Product if
⎟⎠
⎞⎜⎝
⎛ +=3
2sin
3
2cos21
ππiz & ⎟
⎠
⎞⎜⎝
⎛ +=6
11sin
6
11cos82
ππiz
=21zz ⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛ ++⎟⎠
⎞⎜⎝
⎛ +6
11
3
2sin
6
11
3
2cos16
ππππi
⎥⎦
⎤⎢⎣
⎡ +=2
5sin
2
5cos16
ππi [ ] ii 16)1(016 =+=
2
π
Dividing Complex Numbers
( ) ( )[ ]21212
1
2
1 sincos θθθθ −+−= ir
r
z
z
Divide
( )oo iz 300sin300cos241 +=
( )oo iz 75sin75cos82 +=
[ ])75300sin()75300cos(8
24
2
1 oooo iz
z−+−=
[ ]oo i 225sin225cos3 +=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
2
2
2
23 i i
2
23
2
23−−=
2
1
z
z
DeMoivre’s Theorem and nth Roots
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zn = r cosθ + isinθ( )[ ]n
( )θθ ninr n sincos +=
Ex. Find 12)31( i+−
Imagine how much fun it would be to multiply this example out 12 times. It would take forever. Using DeMovre’s Theorem, however, makes it short and simple.
First, convert to trigonometric form.
=+− 12)31( i12
3
2sin
3
2cos2 ⎥
⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛ +ππ
i
( ) ( ) ⎥⎦
⎤⎢⎣
⎡ +=3
212sin
3
212cos212 ππ
i ( )ππ 8sin8cos4096 i+=
( )i014096 += 4096=
nth Roots of a Complex Number
For a positive integer n, the complex number
z = r(cos x + i sin x)
has exactly n distinct nth roots given by
⎟⎠
⎞⎜⎝
⎛ ++
+n
ki
n
krn
πθπθ 2sin
2cos
where k = 0, 1, 2, . . . , n - 1
WATCH THE EXAMPLES CAREFULLY!!!
Ex. Find all the sixth roots of 1.
First, write 1 + 0i in trig. form.
1 + 0i is over 1 and up 0. Therefore, 1 is the hypotenuse and theta is 0o.
( ) 61
01 i+
€
1 cos0o + isin0o( )[ ]1
6
⎟⎠
⎞⎜⎝
⎛ ++
+=
6
20sin
6
20cos16 ππ k
ik
Now, we plug in 0, 1, 2, 3, 4, and 5 for k to find our six roots.
( ) 10sin0cos1 =+ i
ii2
3
2
1
3sin
3cos1 +=⎟
⎠
⎞⎜⎝
⎛ +ππ
ii2
3
2
1
3
2sin
3
2cos1 +−=⎟
⎠
⎞⎜⎝
⎛ +ππ
( ) 1sincos1 −=+ ππ i
ii2
3
2
1
3
4sin
3
4cos1 −−=⎟
⎠
⎞⎜⎝
⎛ +ππ
ii2
3
2
1
3
5sin
3
5cos1 −=⎟
⎠
⎞⎜⎝
⎛ +ππ
Find the three cube roots of z = -2+2i
Again, first convert to trig form.
€
−2 + 2i( )1
3 = 8 cos135o + isin135o( )[ ]1
3
€
81
2 ⎛ ⎝ ⎜ ⎞
⎠ ⎟1
3cos
135 + 360k
3+ isin
135 + 360k
3
⎛
⎝ ⎜
⎞
⎠ ⎟
For k = 0, 1, and 2, the roots are:
( )oo i 45sin45cos2 +
( )oo i 165sin165cos2 +
( )oo i 285sin285cos2 +